### Subject: Pendulum: What is relationship of the T^2 vs. length?

Date: Tue Dec 24 10:02:04 2002
Posted by Michael
Grade level: 10-12 School: Life Sciences Secondary
City: New York State/Province: NY Country: US
Area of science: Physics
ID: 1040742124.Ph
Message:

In a pendulum lab, one gets a straight line if he/she graphs the period-squared (T^2) vs. the length of the string.

The relationship equation is T = 2*pi*SQRT(length/g).

If you graph the string length on the y-axis and the T^2 on the x-axis (which would would technically be incorrect because the string length is the independant variable), the slope's units are meters/sec^2.

Does this slope represent the value of g (acceleration due to gravity)? If so, then how can it? I solved the equation T = 2*pi*SQRT(length/g) for "g."

g = length/(T/2*pi)^2

So how can one simply "exclude" the 2*pi factor and assume that the above slope represents "g." In a lab book, I see a question: "What does the slope represent? [Hint: solve your equation for length/T2.]" My answer would be that it represents nothing of importance. Only graphing "length" vs. "(T/2*pi)^2" would represent "g," which would be important.

Is my thinking off?

MRP

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